SH1009, Modern Fysik. X-Ray Diffraction Muhammad Yasir, Lway al Maeeni, Joakim Wahlström

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1 SH1009, Modern Fysik X-Ray Diffraction Muhammad Yasir, Lway al Maeeni, Joakim Wahlström 24 th april, 2009

2 Introduction In this lab we study the x-rays and their ability to penetrate materials. We also study the variation of x-ray intensities and its dependence on different materials. When a charged particle is moving in an accelerated motion it is continuously emitting electromagnetic radiation. This process occurs very frequently in nature and gives rise to the well known electromagnetic spectrum which extends sixteen orders of magnitude in energy or wavelength. The electromagnetic spectrum only describes the continuous radiation emitted from charged particles in an accelerated motion. In addition discrete radiation is emitted over the same energy range. Discrete x-rays are emitted at transitions between inner atomic shells. Experimentation Equipment 1. Crystal spectrometer, 2. Aluminum, zinc, copper, nickel and tin samples of different thickness. 3. Computer program mesh Procedure We started by placing the LiF crystal in the crystal spectrometer and record the x-ray spectrum. The voltage applied was 25keV. K α and K β radiations were observed along with their corresponding angles. With the help of Braggs law wavelengths associated with these radiations were determined. The spectrometer arm was then adjusted to the angle corresponding to where K β radiation had its maximum. Aluminum with 5 different thicknesses was placed in front of the detector and graphs of intensity as function of thickness x were produced. Five different values for mass absorption coefficient µ were calculated with the help of the following formula: x ( ) = I x I e µ o The different values of µ were plotted with thickness on the x axis and an exponential fit was used to determine its value. The mass coefficients of nickel, zinc, copper and tin were determined in the same fashion but only using a constant thickness of each material. The values were plotted in a graph with atomic number Z on x axis.

3 Calculation Bragg diffraction formula 2 For Lithium florid (LiF) it was given d = 201 pm. After the experimentation two peaks corresponding to the first order (n=1) were found: 19,9 22,2 From which the corresponding wavelengths were calculated 1, , Now a KBr crystal is used, to the same experimentation, and clear peaks of order n=1 and n=2 are shown: 1: 11,9 13,3

4 2: 24,7 27,5 For KBr we calculate the value of d by using the known λ and the Bragg equation for n=1: 2 So with both wavelengths we get an average of 330, which is very close to the real value that is written on the crystal (d=329 pm). Now the aim is to find the value of µ for different elements by using the formula:. Firstly I 0 was decided by running the experiment without putting any material on the anode and fixing the intensity for the angle for K β (i.e. 19,9 ), I 0 =707. By putting aluminum with different thicknesses: X I µ(al) 0, , , ,3367 0, ,1676 0, ,0371 0, ,49743

5 We could then plot the intensity I as a function of the width x and fit an exponential curve to it. 700 Exponential fit for Al f = a*exp(-b*x) 600 I (counts) I= 711*exp(-10032*x) , , , , , , ,00012 x (m) If we compare the given relation, I= 711*exp(-10032*x), with we can conclude that should be around m -1 or 100,32 cm -1. If we want the mass absorption coefficient we have to divide by the density:,,, 37,2 Since we have already calculated the wavelength corresponding to the angle used we can obtain the energy from: 9,06, A plot from NIST reveals that this is quite close to the actual value.

6 Later for Ni, Cu, Zn and Sn, all of them with the same thickness (x=0,0250 mm) Ni: I= ,7 Cu: I= ,94 Zn: I= ,56 Sn: I= Z gram/cm - 3 µ m (cm 2 /gram) µ (m -1 ) I (counts) Ni 28 8, , ,7 16 Cu 29 8,94 38, , Zn 30 7,14 40, , Sn 50 7, , ,1 28 If we plot µ against the atomic number Z we end up with the following strange looking plot. There doesn t seem to be anything particular that relates the mass absorption coefficient with the atomic number.

7 μ m (cm 2 /gram) Atomic number Z If we instead take values from the plots at NIST we get the following. Z µ (m -1 ) Ni Cu Zn Sn We notice that the last three values seem to agree with our measurements while the first one is way too high. A plot of the values from NIST for Ni, Cu and Zn seem to give a linear relationship μ (m -1 ) , , , ,5 Atomic number Z Preparatory question 3: We are asked to calculate the width of a specific material at which 99% of the radiation is prohibited from going through. Some numerical values are given (for 50 kev x-rays) and the relation by which this phenomenon can be described, namely: 0.

8 Solving the above equation for x (width) give us: In order for this equation to give a unit of thickness we have to multiply the mass absorption coefficient with the density. Then knowing the ratio of 0 versus as, we can easily calculate the required width.,, 0 1 ln 100 0,01 C 0,18 2,2 0, ,2922 Al 0,38 2,7 1,026 44,8847 Fe 2 7,9 15,8 2, Sn 11 7,3 80,3 0, Conclusion We can conclude that even though Bragg s initial assumptions are elementary his diffraction formula works very well in describing the constructive interference of x-rays in crystals. The values for the mass absorption coefficient did correspond fairly well with the values from NIST in four out of five cases, with the value for nickel being too high. We can only speculate as to why we got this result, but it was most probably due to contamination of the nickel-absorber.

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